Problem: $\sum\limits_{n=1}^{\infty } \dfrac{-2(x+6)^n}{n\cdot(-5)^n} $ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $-11<x \le -1$ (Choice B) B $-11 \le x < -1$ (Choice C) C $-1 \le x < 11$ (Choice D) D $-1<x \le 11$
We use the ratio test. For $x\neq-6$, let $a_n= \dfrac{-2(x+6)^n}{n\cdot(-5)^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|= \left| \dfrac{x+6}{5}\right|$ The series converges when $\left| \dfrac{x+6}{5}\right|<1$, which is when $-11<x<-1$. Now let's check the endpoints, $x=-11$ and $x=-1$. Letting $x=-11$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{-2(-11+6)^n}{n\cdot(-5)^n} &= \sum\limits_{n=1}^{\infty } \dfrac{-2(-5)^n}{n\cdot(-5)^n} \\\\ &= -2\sum\limits_{n=1}^{\infty } \dfrac{1}{n} \end{aligned}$ This is a multiple of the harmonic series, which is known to diverge. Letting $x=-1$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty } \dfrac{-2(-1+6)^n}{n\cdot(-5)^n} &= \sum\limits_{n=1}^{\infty } \dfrac{-2(5)^n}{n\cdot(-5)^n} \\\\ &= \sum\limits_{n=1}^{\infty } \dfrac{-2(-1)^n(-5)^n}{n\cdot(-5)^n} \\\\&= -2\sum\limits_{n=1}^{\infty } \dfrac{(-1)^n}{n} \end{aligned}$ This is a multiple of the alternating harmonic series, which is known to converge. In conclusion, the interval of convergence is $-11<x \le -1$.